Existence of matrix $C$ such that $f(A)=tr(AC)$

Question

Let $f: M_{n\times n}(\mathbb R)\to \mathbb R$ a functional. Prove that there exists a unique square matrix $C$ such that $f(A)=tr(AC)$ for all $A\in M_{n\times n}(\mathbb R)$

I´ve been trying to solve this problem but unfortunately I don´t know where to start. I would really appreciate if you can give me some hints or suggestions

If the functional is continuous and linear, the Riesz representation theorem assures the unique existence of matrix $C$ by considering the norm $|A| = \sqrt{\operatorname{tr}(AA^t)}$. — corindo, Nov 09 '15 at 01:41
Would you know how to do this for an $f:\Bbb R^{n^2}\to \Bbb R$? — Ben Grossmann, Nov 09 '15 at 01:51

score 0 · Answer 1 · answered Nov 09 '15 at 01:52

0

Hint: Note that $\operatorname{trace}(AB^T)$ is just the "dot product" of $A$ and $B$.

answered Nov 09 '15 at 01:52

Ben Grossmann

225,327

Thanks for the hint!. I´ll try to solve it – user128422 Nov 09 '15 at 02:04

Existence of matrix $C$ such that $f(A)=tr(AC)$

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